Matrix Multiplication

Question

If I had any $m\times n$-matrix, where $m$ and $n$ are variables, what can I multiply this matrix by?

Can it be multiplied by itself?

By a square matrix?

Another $m\times n$-matrix?

Or can it be multiplied by a scalar?

To be complete, there are also less common ways to multiply matrices, that aren't part of @Cameron's excellent answer. — vadim123, Jan 14 '14 at 00:39

score 2 · Accepted Answer · answered Jan 14 '14 at 00:22

2

It can be multiplied componentwise by a scalar. It can be multiplied on the right by any $n$-row matrix, and on the left by any $m$-column matrix.

answered Jan 14 '14 at 00:22

Cameron Buie

102,994

1

So it could be by a scalar? That's what I was thinking. – PrecalcNewb Jan 14 '14 at 00:25
Yes, we can also multiply it by a scalar, though in general matrix multiplication and scalar multiplication are not the same thing. I would make a note that $m$ and $n$ should not be variables (as the dimensions of a particular matrix cannot change). Rather, they are some arbitrary fixed values (positive integers, specifically). – Cameron Buie Jan 14 '14 at 00:32
That makes more sense, thank you. – PrecalcNewb Jan 14 '14 at 00:33

user127.0.0.1 · Answer 2 · 2014-01-14T00:29:28.857

1

$A\cdot B$ is defined, iff $A$ has as many columns as $B$ has rows

So, if $A$ is a $n\times m$-matrix and $B$ is a $m\times p$-matrix the product $A\cdot B$ is well-defined and the result is a $n\times p$-matrix.

Moreover you are allowd to multiply any matrix by a given scalar.

edited Jan 14 '14 at 00:29

answered Jan 14 '14 at 00:23

user127.0.0.1

7,267

Fantastic, thank you all. – PrecalcNewb Jan 14 '14 at 00:32

score 0 · Answer 3 · answered Jul 16 '15 at 11:51

Just to add a little to the answers here... there are a few details missing here but this is a good reference for the details.

If the entries of your $m\times n$ matrix $A=[a_{ij}]$ lie in a ring $R$ (you can think $\mathbb{R}$ or $\mathbb{C}$ probably) then your matrix can be seen as a linear map, $T_A$: $$T_A:R^n\rightarrow R^m\,,\,e_j\mapsto \sum_ia_{ij}e_i,$$

where $a_{ij}$ is the $(i,j)$-entry of $A$. That is $T_A$ takes in $n$-scalars and 'spits' out $m$.

You can show that the composition of linear maps as matrices is given by the ordinary matrix multiplication.

Now you can compose (multiply) with a linear map $T_B$ (matrix $B$) on the left or the right. Consider first on the left: $BA$. Now this is a linear map given by $T_B\circ T_A$:

$$R^n\underset{T_A}{\longrightarrow} R^m\underset{T_B}{\longrightarrow} V,$$

where $V$ is some vector space. If $B$ is to be a matrix with entries from $R$ then $V=R^k$ for some $k\in\mathbb{N}$.

Therefore we have:

$$R^n\underset{T_A}{\longrightarrow} R^m\underset{T_B}{\longrightarrow} R^k,$$

and so $A$ can be multiplied on the left only by a matrix $B$ that induces a linear map

$$T_B:R^m\rightarrow R^k;$$

i.e. $B$ is a $k\times m$ matrix. Therefore to multiply $B$ by $A$ on the left we have

$$\underbrace{B}_{k\times m}\quad\cdot\quad\underbrace{A}_{m\times n};$$

i.e. what the other answers said. Note that $T_B\circ T_A:R^n\rightarrow R^k$ so that $BA$ is a $k\times n$ matrix.

You can show that for multiplication on the right you need an $n\times k$ matrix.

Matrix Multiplication

3 Answers3